$\dfrac{dy}{dx}=3y$ and $y(0)=3$. $y(\ln 2)=$
The differential equation is separable. What does it look like after we separate the variables? $\dfrac{dy}y=3\,dx$ Let's integrate both sides of the equation. $\int\dfrac{dy}y=\int3\,dx$ What do we get? $\ln |y| = 3x+C$ What value of $C$ makes the solution curve pass through the point $(0,3)$ ? Let's substitute $x=0$ and $y=3$ into the equation and solve for $C$. $\begin{aligned} \ln |3| &= 3\cdot0+C\\ \\ C&=\ln 3 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\ln |y| = 3x+\ln 3$ $|y| = e^{3x+\ln 3}$ $|y| = e^{3x}\cdot e^{\ln 3}$ $|y| = 3e^{3x}$ $y = \pm3e^{3x}$ Which sign makes $y=3$ when $x=0$ ? We must choose the positive sign to pass the solution curve through the point $(0,3)$. $ y=3 e^{3x}$ What is $y$ when $x=\ln 2$ ? $\begin{aligned} y&=3e^{3\ln 2}\\ \\ &=3e^{\ln 8}\\ \\ &=3\cdot8\\ \\ &=24 \end{aligned}$